Solving 2D Kinematics Problem: Distance OP in m

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The problem involves an arrow shot horizontally towards point O, which is 20.0 m away, hitting point P below O after 0.10 seconds. The key to solving the problem lies in understanding that the vertical motion of the arrow is influenced by gravity, with an acceleration of 9.8 m/s². Since the initial vertical velocity is zero, the distance OP can be calculated using the formula for free fall: x = 0.5*a*t². By substituting the values, the distance OP can be determined. This scenario emphasizes the importance of analyzing kinematics in one dimension, particularly in the vertical direction.
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I have a problem with this question.
An arrow is shot horizontally towards point O, which is at a distance of 20.0 m. It hits point P (right below point O) 0.10 s later. What is the distance OP (in metres)?
What do i need to find? How would i know the time taken for the arrow to reach o?
 
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The distance to the target is a red herring; this is a question of kinematics in 1D: the vertical dimension.

You know the arrow has been in free fall for 0.10s, and when something is in free-fall it accelerates downwards at 9.8 m/s^2. Since the arrow was shot horizontally, it's initial downward velocity is zero, and you can use

x = 0.5*a*t^2 + v*t

to find your distance.
 

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